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In the following figure, If MN ll BC, MN devides the triangle into two equal parts, then the value of ratio of MA and AB will be ?
A) sqrt2
B) 1/sqrt2
C) sqrt(2+1)/sqrt2
D) sqrt(2-1)/sqrt2
E) sqrt(2+1)/2
Written by Take GRE Team on July 21st, 2008 with
16 comments.
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In the following figure, If MN ll BC, MN devides the triangle into two equal parts, then the value of ratio of MA and AB will be ?
A) sqrt2
B) 1/sqrt2
C) sqrt(2+1)/sqrt2
D) sqrt(2-1)/sqrt2
E) sqrt(2+1)/2
GRE Question of the Day by Take GRE Team on July 21st, 2008 with
16 answers.
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#2.
July 18th, 2006, at 12:51 PM.
It’s very “…………..” we dont have word to fill the blank.
Hey 1600 Crackers !! This question will clarify one of the important aspect of geometry. Which normally not mentioned, in deatails, in any GRE prep books.
niharika
#3.
July 19th, 2006, at 10:32 AM.
i think its ..A.admin.!! can u plzzz give the answer as i have my exam veryyyyy soon??
swapnil
#4.
July 19th, 2006, at 2:10 PM.
hey whats sqrt? what does it stand for? it doesnot appear in the picture
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#5.
July 19th, 2006, at 2:12 PM.
It is writing form of square root.
For Example:
sqrt4=2
swapnil
#6.
July 19th, 2006, at 2:31 PM.
MN is parallel to BC , MN divides triangle ABC into two equal parts, From this information can we say that MN = 1/2 BC, by midpoint theorem ? If yes then it can help us arrive at the answer.
veenit
#7.
August 6th, 2006, at 11:44 PM.
ma=2ab…..what else..no idea..completely absurd
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#8.
August 7th, 2006, at 7:17 AM.
hmm no one with clear thoughts.
Abhijit
#9.
August 16th, 2006, at 5:43 PM.
b) i think its 1/sqrt2
if i am not wrong….
triangle ABC and triangle AMN are similar triangles
hence MA^2/AB^2 = Area(AMN)/Area(ABC)……(property of similar triangles)
Area(ABC) = 2 * Area(AMN)
so MA/AB = 1/sqrt2
…is this correct …
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#10.
August 16th, 2006, at 6:02 PM.
Yeah abhijit that is correct choice.
subbu.
#11.
November 20th, 2006, at 8:19 AM.
who said that area of a triangle is equal to square of any side.. even we cannot take it proportional also… wat will u do if it is a right triangle..???????
bhavinsurela
#12.
July 21st, 2008, at 12:21 PM.
i dnt think.. this is d property!..
prosenjit konar
#13.
July 21st, 2008, at 9:14 PM.
take gre team cn u xpln….i suppose there is another info whch s nt provided
Allen
#14.
July 22nd, 2008, at 8:42 AM.
Assume an equilateral triangle. It is a missing detail.
Narayan
#15.
July 23rd, 2008, at 9:04 PM.
b) it should be 1/sqrt2
AMN and ABC are similar triangles. Their area will be proportional to the squares of their sides. The are of ABC will be square of AB/AM x area of AMN i.e. double, which agrees with the given information that MN devides the triangle into two equal parts.
naungsai
#16.
July 28th, 2008, at 11:58 AM.
I think Abhijit is correct.
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#1. July 17th, 2006, at 5:11 PM.
plz…help us out with the solution..??